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CHEMISTRY XL-14A CHEMICAL EQUILIBRIA August 20, 2011 Robert Iafe Unit Overview 2 Reactions at Equilibrium Equilibrium Calculations Le Châtelier’s Principle Catalysts Reactions at Equilibrium 3 Reversibility of Reactions Law of Mass Action Gas Phase Equilibrium Solution Phase Equilibrium Extent of Reaction Direction of Reaction K KP KC using K Q vs K Introduction to Chemical Equilibrium 4 We previously assumed all reactions proceeded to completion However: many reactions approach a state of equilibrium Equilibrium – condition of a chemical reaction in which chemical change ceases and no further change occurs spontaneously Equilibrium – a dynamic equilibrium between reactants and products in a chemical reaction. At Equilibrium: Forward and reverse reactions simultaneous with equal rates No further net conversion of reactants to products unless the experimental conditions have been changed The equilibrium state is characterized by the Equilibrium Constant (Keq) Reversibility of Reactions 5 A. N2 (g) + 3 H2 (g) 2 NH3 (g) B. 2 NH3 (g) N2 (g) + 3 H2 (g) A B Reversibility of Reactions 6 Dissolve CoCl2.6H2O in pure water Dissolve in 10 M HCl Add conc. HCl Add H 2O [Co(H2O)6]2+ [Co(H2O)6]2++ [CoCl4]2- [CoCl4]2- Pink Purple Blue [Co(H2O)6]2+ + 4 Cl- [CoCl4]2- + 6 H2O The Nature of Chemical Equilibrium 7 [Co(H2O)6]2+ + 4 Cl- [CoCl4]2- + 6 H2O %Co 100 98 [Co(H2O)6]2+ A [Co(H2O)6]2+ A DC1 2 [CoCl4]2- B DC2 [CoCl4]2- Dt1 Dt2 As time progresses, rate = concentration/time slows When rate(s) = constant, reaction is at equilibrium B The Nature of Chemical Equilibrium Characteristics of the Equilibrium State 8 1. No Macroscopic Change 2. Reached Spontaneously 3. Dynamic Balance of Forward/Reverse Processes 4. Same regardless of direction of approach [Co(H2O)6]2+ + 4 Cl- [CoCl4]2- + 6 H2O Situations which appear to be equilibrium, but are not: Steady State – macroscopic concentrations are constant, even though system is not at equilibrium 1 process removes species while 2nd process supplies species Homeostasis – body tries to maintain blood pH, etc… Equilibrium Data 9 1864: Norwegians Cato Guldberg and Peter Waage discovered the mathematical relationship that summarized the composition of a reaction mixture at equilibrium 2 SO2 (g) + O2 (g) P P D B 2 SO3 (g) at 1000. K P P P P d E b C P P e c K The Empirical Law of Mass Action 10 bB + cC dD + eE At equilibrium (independent of starting conditions) the law of mass action is constant d D b B e E c C aa K aa (unitless) aB = ‘activity’ of species B K = equilibrium constant Magnitude of K tells us about the equilibrium state K >> 1, (activities of products) >> (activities of reactants) at equilibrium K << 1, (activities of products) << (activities of reactants) at equilibrium ‘Activity’ 11 Ideal Gas d D b B e E c C aa K aa PB aB P P1 ba Ideal Solution [B] aB c c 1M Pure Liquid a=1 Pure Solid a=1 Gas Phase Reactions 12 bB + cC d D b B e E c C aa K aa P P D B P P P P d E b C d e D E b c B C a = activity of reacting species For ideal gases: P P P KP P P P dD + eE PB aB P e c d e D E b c B C P P K P P P K KP K P where Pº = 1 bar d ebc d ebc ****Note**** Your textbook refers to KP as K Gas Phase Reactions 13 Write the equilibrium expression for the following reaction: CO(g) + ½ O2(g) CO2 (g) Molarity 14 Molarity is a unit commonly used to describe the concentration of a solution Solution – homogeneous mixture of two or more substances Solute – one or more minor components in a solution Solvent – the major component of a solution, medium for solute Molarity – the number of moles per unit volume (mol/L = M) n mol solute Molarity M V L solution It can also be used to describe the molar volume of a gas n mol gas Molarity M V L gas Reactions in Solution 15 bB + cC d D b B e E c C aa K aa a = activity of reacting species [B] For ideal solutions: aB where cº = 1 M c [D] c [E] c [B] c [C] c d e b c [D]d [E]e b dD + eE c [B] [C] K KC [D]d [E]e b c [B] [C] K c KC K c d ebc d ebc Reactions in Solution 16 Write the equilibrium expression for the following reaction: Cl2(aq) + 2OH-(aq) H2O(l) ClO-(aq) + Cl-(aq) + Pure Substances and Multiples Phases 17 d D b B e E c C aa K aa H2O(l) I2(s) CaCO3(s) a = activity of reacting species For solids: a(pure solid) = 1 For liquids: a(pure liquid) = 1 H2O(g) K P H O 2 I2(aq) I2 K CaO(s) + CO2(g) K P CO 2 Mixed Phase Reaction 18 Write the equilibrium expression for the following reaction: Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l) 2 H O 2 (l) a a 2 a H Zn (aq ) aq 2 K 2 a aO Zn (s) H 3 aq K Zn 2 /1M PH 2 /1atm1 1H 3O 2 /1M 2 Zn P K H O 2 H2 2 3 Extent of a Reaction 19 Phosgene (COCl2) is an important intermediate in the manufacture of certain plastics. It is produced by the reaction: CO(g) + Cl2(g) a) b) COCl2(g) Use the law of mass action to write the equilibrium expression for this reaction At 600 C, K = 0.20. Calculate the partial pressure of phosgene in equilibrium with a mixture of PCO = 0.0020 atm and PCl2 = 0.00030 atm Direction of Change in Reactions: Reaction Quotient (Q) 20 To determine the direction of a reaction, use the Reaction Quotient: d e D E b c B C P P Q P P D E Q b c B C d e Q is the same expression as K Q valid all the time K valid only at equilibrium The relationship between Q and K will determine the direction of a reaction Direction of Change in Reactions: Q vs K 21 When K < Q: K<Q Reaction proceeds in reverse (left) (too little reactant, too much product When Q = K: Q K=Q Reaction is at equilibrium K>Q When K > Q: Reaction proceeds forward (right) (too much reactant, not enough product) time Equilibrium Calculations 22 K vs KC Relationships between K’s of reactions Using K and Kc ICE box calculations K vs KC 23 [D]deq [E]eeq b eq c eq [B] [C] d e PD,eq PE,eq KC nX RT PX V nX X V b c B,eq C,eq P P K L bar R 8.3145 10 mol K 2 PX X RT K vs KC 24 d e D,eq E,eq b c B,eq C,eq P P P P K PX X RT X D RT X E RT P P K b c P P X B RT XC RT d d e D E b c B C X D X E d ebc d ebc K RT K c RT b c X B XC d e e K K c RT d ebc L bar R 8.3145 10 mol K 2 K vs KC 25 At 1132 ºC, for the following reaction: 2 H2S(g) 2 H2(g) + S2(g) KC = 2.26 x 10-4. What is the value of K at this temperature? K K c RT d ebc L bar R 8.3145 10 mol K 2 Relationships among Equilibrium Expressions 26 Case 1: Compare forward and reverse reactions Case 2: Multiply a chemical reaction by a constant x Case 3: Add or subtract chemical reactions (Hess’s Law) 2 H2(g) + O2(g) K1 2 H2O(g) 2 H2O(g) 2 H2(g) + O2(g) K1K1' PH2 2O PH2 2 PO2 2 H2 2 H 2O P PO2 P 1 K1' PH2 2O PH2 2 PO2 PH2 2 PO2 PH2 2O 1 K1 K1' Relationships among Equilibrium Expressions 27 Case 1: Compare forward and reverse reactions Case 2: Multiply a chemical reaction by a constant x Case 3: Add or subtract chemical reactions (Hess’s Law) 2 H2(g) + O2(g) 2 H2O(g) H2(g) + ½ O2(g) H2O(g) K1' PH2 2O 2 H2 P PO2 K1 K1 K1' K1' K x 1 PH2 2O PH2 2 PO2 PH 2O PH 2 PO1/2 2 Relationships among Equilibrium Expressions 28 Case 1: Compare forward and reverse reactions Case 2: Multiply a chemical reaction by a constant x Case 3: Add or subtract chemical reactions (Hess’s Law) 2 BrCl(g) Cl2(g) + Br2(g) Br2(g) + I2(g) 2 IBr(g) 2 BrCl(g) + I2(g) K3 2 PBrCl 2 PIBr K2 PBr2 PI 2 2 IBr(g) + Cl2(g) 2 PIBr PCl 2 2 BrCl I 2 P K1 PCl 2 PBr2 P PCl 2 PBr 2 2 BrCl P 2 IBr P K K 1 2 PI 2 PBr2 1+2 K3 = K1 x K2 If subtracting reactions: 1 – 2 K3 = K1 / K2 If adding reactions: K3 2 PIBr PCl 2 2 PBrCl PI 2 Relationships among Equilibrium Expressions 29 At 1330 K, germanium(II) oxide (GeO) and tungsten(VI) oxide (W2O6) are both gases. The following two equilibria are established simultaneously: 2GeO(g) + W2O6(g) 2 GeWO4(g) K = 7000 GeO(g) + W2O6(g) GeW2O7(g) K = 38,000 2 GeWO4(g) K=? Compute K for the reaction: GeO(g) + GeW2O7(g) Equilibrium Calculations: (R)ICE Box 30 For a Chemical Reaction: bB + cC bB Initial Pa 1 atm 1 atm 0 atm Change in Pa -bx -cx + dx Equilibrium Pa 1 - bx 1 - cx 0 + dx P P K P P cC eE Reaction d e D E b c B C + dD + dD + eE 0 atm +ex 0 + ex dx ex K b c 1 bx 1 cx d e Evaluating K from Reaction Data 31 At 600 ºC, a gas mixture of CO and Cl2 has initial partial pressures of P(CO) = 0.60 atm and P(Cl2) = 1.10 atm. At equilibrium, the P(COCl2) = 0.10 atm. Calculate K for this reaction: CO(g) Initial Pa Change in Pa 0.60 atm --x0.10 + Cl2(g) 1.10 atm -x - 0.10 (1.10-0.10) ? atm - x atm 0.60? atm - atm x 1.10 Equilibrium Pa (0.60-0.10) K PCOCl2 PCO PCl 2 COCl2(g) 0 atm x ++0.10 0.10 atm 0.10 0.10 K 0.20 0.60 0.101.10 0.10 0.501.00 Evaluating K from Reaction Data 32 Must take into account stoichiometric coefficients: 2 As(OH)63-(aq) + 6 CO2(g) As2O3(s) + 6 HCO3-(aq) + 3 H2O(l) 2 As(OH)63-(aq) + 6 CO2(g) I [A] 1M 1M C in [A] -2x -6x E [A] KC - HCO 1M - +6x 1 M – 2x 1 M – 6x 6 3 3 2 6 As2O3(s) + 6 HCO3-(aq) + 3 H2O(l) 1 M + 6x 1 6x KC 2 6 1 2x 1 6x 6 As(OH) CO 6 2 Evaluating K from Reaction Data 33 Graphite (solid carbon) is added to a vessel that contains CO2(g) at a pressure of 0.824 atm. The pressure rises during the reaction. The total pressure at equilibrium is 1.366 atm. Calculate the equilibrium constant: C(s) + CO2(g) 2 CO (g) Calculating Equilibrium Compositions 34 H2(g) and I2(g) are sealed in a flask with partial pressures: P(H2) = 1.980 atm and P(I2) = 1.710 atm. The sealed flask is heated to 600 K, and the gases quickly reach equilibrium. K600 K = 92.6 H2(g) + I2(g) 2 HI(g) Calculate the equilibrium partial pressures of H2, I2 and HI at 600 K Can solve this 3 ways: 1) Solve exactly using quadratic equation 2) Solve by approximation 3) Solve by successive approximation if 2 doesn’t work b b 4ac x 2a 2 ax2 + bx + c = 0 Le Châtelier’s Principle 35 Adding and Removing Reagents Changing Pressure Effect of Temperature on Equilibrium Effect of Temperature on K Catalysts Direction of Change in Reactions: Le Châtelier’s Principle 36 Le Châtelier’s Principle: A system in equilibrium that is subjected to a stress will react in a way that tends to counteract the stress At equilibrium, the macroscopic properties of a system remain constant When you perturb the equilibrium, the system will counteract the perturbation • Changing the concentration of a reactant or product • Changing the Volume or Pressure • Changing the Temperature Can maximize the yield by perturbing the system Le Châtelier’s Principle Adding or Removing Reagent 37 Change the concentration (or Partial Pressure) of a reagent Consider how the change in concentration affects Q H2(g) + I2(g) Q PH 2 PI 2 2 HI P Increase a Product (decrease a Reactant) Q>K Reverse (to the left) Increase a Reactant (decrease a Product) Q<K PHI 3.0 Forward (to the right) PX (atm) 2HI(g) 2.0 1.0 PH 2 PI 2 Le Châtelier’s Principle Changing Volume or Pressure 38 Specific to gas phase reactions 2 P2(g) P4(g) Increase V (decrease P) Reaction direction with greater ngas Increase P (decrease V) Reaction direction with fewer ngas Le Châtelier’s Principle Changing the Temperature 39 Endothermic or Exothermic reactions Endothermic Reaction – absorbs heat Exothermic Reaction – gives off heat Increase T (“adding heat”) Reaction will proceed endothermically Trying to absorb excess heat Increase T Endothermic reaction forward Exothermic reaction reverse Decrease T Endothermic reaction reverse Exothermic reaction forward 2 NO2(g) N2O4(g) H < 0 exothermic Direction of Change in Reactions: Le Châtelier’s Principle 40 3 Al2Cl6(g) 2 Al3Cl9(g) Increase P(Al2Cl6) Increase Volume Decrease Temperature (assume H < 0) H2S(g) + l2(g) 2 HI(g) + S(s) The equilibrium constant at 110 ºC is K = 0.0023 For the following conditions, calculate Q and determine the direction of the reaction: a) P(I2) = 0.461 atm, P(H2S) = 0.050 atm, P(HI) = 0.1 atm b) P(I2) = 0.461 atm, P(H2S) = 0.050 atm, P(HI) = 1.0 x 10-3 atm Temperature Dependence of K 41 Van’t Hoff Equation K2 DH 1 1 ln K1 R T2 T1 Hº is the Enthalpy of a Reaction Hº < 0 – Exothermic Hº > 0 – Endothermic R = 8.31451 J/mol K Temperature Dependence of K 42 K2 DH 1 1 ln K1 R T2 T1 If H < 0 (exothermic reaction), then Increase in T reduces K If H > 0 (endothermic reaction), then Increase in T increases K Direction of Change in Reactions: van’t Hoff Equation 43 1 1 K D H 2 ln K R T T 1 2 1 3 H2(g) + N2(g) 2 NH3(g) K(298 K) = 5.9 x 105 If Hº = -92.2 kJ/mol, calculate K at T = 600 K Catalysts 44 A Catalyst is a substance that increases the rate of a chemical reaction and is not consumed in the net reaction 2 K Haber Process: N2(g) + 3 H2(g) Fe3O4 2 NH3(g) PNH 3 PN 2 PH3 2 ΔHº = -92.4 kJ/mol composition Catalysts do not affect the equilibrium They only change the path Catalysts speed up both the forward and reverse reactions You will learn more about catalysts in 14B when you study Reaction Kinetics